What is the `Delta`H of reaction in kJ/mol of n-C6H14 for the reaction described in the image?

The heat energy produced by the combustion of 1.00g of n-hexane is equal to the  sum of the heat absorbed by the water and the heat absorbed by the calorimeter.  The specific heat of water tells you how much heat is absorbed in raising the temperature of one gram one degree Celsius. The water equivalent of the calorimeter is the mass of water that absorbs the same amount of heat as the calorimeter.

The calorimeter and water reach thermal equilibrium, so both have a `Delta`T of (29.30-22.64)=6.66 degrees C.

Heat gained by water: q= mc`Delta`T = (1500g)(4.18 J/g-degC)(6.66 degC)=41,758 J

Heat gained by calorimeter: q= mc`Delta`T = (961g)(4.18J/g-degC)(6.66 degC) =26,753 J

The total heat absorbed: q = 41,758 J + 26,753 J = 68,511 J = 68.5 kJ

The heat released by the combustion of n-hexane is opposite in sign to that gained by the water and calorimeter, so heat of the reaction is q = -68.5 kJ

Now to find the kJ per mole of n-hexane we need to calculate the number of moles that reacted:

1.00 g x 1 mol/86.1g = 0.0116 moles

`Delta` H= q/moles = -68.5 kJ/0.0116 mol = 5.91 x 10^3 kJ/mol